# Problem of the Week Problem D and Solution That Triangle

In the diagram, $$ABCD$$ is a rectangle. Point $$E$$ is outside the rectangle so that $$\triangle AED$$ is an isosceles right-angled triangle with hypotenuse $$AD$$. Point $$F$$ is the midpoint of $$AD$$, and $$EF$$ is perpendicular to $$AD$$.

If $$BC=4$$ and $$AB=3$$, determine the area of $$\triangle EBD$$.

## Solution

Since $$ABCD$$ is a rectangle then $$AD=BC=4$$. Since $$F$$ is the midpoint of $$AD$$, then $$AF=FD= 2.$$

Since $$\triangle AED$$ is an isosceles right-angled triangle, then $$\angle EAD = 45^\circ$$.
Now in $$\triangle EAF$$, $$\angle EAF = \angle EAD = 45^\circ$$ and $$\angle AFE= 90^\circ$$.
Since the sum of the angles in a triangle is $$180^\circ$$, then $$\angle AEF = 180^\circ - 90^\circ - 45^\circ = 45^\circ$$.
Therefore, $$\triangle EAF$$ has two equal angles and is therefore an isosceles right-angled triangle.
Therefore, $$EF=AF = 2$$.

From this point we are going to look at two different solutions.

Solution 1:

We calculate the area of $$\triangle EBD$$ by adding the areas of $$\triangle BAD$$ and $$\triangle AED$$ and subtracting the area of $$\triangle ABE$$.
Since $$AB=3$$, $$DA=4$$, and $$\angle DAB= 90^\circ$$, then the area of $$\triangle BAD$$ is $$\frac{1}{2} (3)(4) = 6$$.
Since $$AD = 4$$, $$EF = 2$$, and $$EF$$ is perpendicular $$AD$$, then the area of $$\triangle AED$$ is $$\frac{1}{2} (4)(2) = 4$$.
When we look at $$\triangle ABE$$ with the base being $$AB$$, then its height is the length of $$AF$$.
Therefore, the area of $$\triangle ABE$$ is $$\frac{1}{2} (3)(2)=3$$.

Therefore, the area of $$\triangle EBD$$ is $$6 + 4 - 3 = 7$$.

Solution 2:

Extend $$BA$$ to $$G$$ and $$CD$$ to $$H$$ so that $$GH$$ is perpendicular to each $$GB$$ and $$HC$$ and so that $$GH$$ passes through $$E$$.
Each of $$GAFE$$ and $$EFDH$$ has three right angles (at $$G$$, $$A$$, and $$F$$, and $$F$$, $$D$$, and $$H$$, respectively), so each of these is a rectangle.
Since $$AF= EF = FD = 2$$, then each of $$GAFE$$ and $$EFDH$$ is a square with side length 2.
Now $$GBCH$$ is a rectangle with $$GB = GA + AB = 2 + 3 = 5$$ and $$BC = 4$$.

The area of $$\triangle EBD$$ is equal to the area of rectangle $$GBCH$$ minus the areas of $$\triangle EGB$$, $$\triangle BCD$$, and $$\triangle DHE$$.

Rectangle $$GBCH$$ is 5 by 4, and so has area $$5 \times 4 = 20$$.
Since $$EG = 2$$ and $$GB = 5$$ and $$EG$$ is perpendicular to $$GB$$,
then the area of $$\triangle EGB$$ is $$\frac{1}{2} (EG)(GB) = \frac{1}{2}(2)(5)=5$$.
Since $$BC = 4$$ and $$CD=3$$ and $$BC$$ perpendicular to $$CD$$,
then the area of $$\triangle BCD$$ is $$\frac{1}{2} (BC)(CD) = \frac{1}{2}(4)(3)=6$$.
Since $$DH = HE = 2$$ and $$DH$$ is perpendicular to $$EH$$,
then the area of $$\triangle DHE$$ is $$\frac{1}{2} (DH)(HE)=\frac{1}{2}(2)(2)=2$$.

Therefore, the area of $$\triangle EBH$$ is $$20 - 5 - 6 - 2 = 7$$.